The request
Give a self-contained Bayesian analysis of a fictional museum's color-urn…
Read the full request
Give a self-contained Bayesian analysis of a fictional museum's color-urn exhibit. This is a mathematical explanation, not software. Return the analysis directly in your final response; do not create or modify files, write scripts, start servers, or use the internet.
An urn is selected once and then kept for all draws. Urn A is selected with probability 1/2 and contains 4 red and 2 blue balls. Urn B is selected with probability 1/3 and contains 2 red and 4 blue balls. Urn C is selected with probability 1/6 and contains 1 red and 5 blue balls. Two balls are drawn without replacement. We are told only that exactly one was red and one was blue; their order is not observed.
Derive a table showing each urn's prior, likelihood of this unordered observation, unnormalized posterior weight, and normalized posterior. Show the evidence probability and verify that the posterior sums to one. Then compute the probability that a third draw from the same urn is red, still without replacement, by explicitly accounting for the two removed balls.
Finally audit a tempting but wrong shortcut: replacing the model with independent draws with replacement. Compute that alternative model's evidence probability, posterior, and third-red prediction. State precisely which quantities coincide and which differ, explain why a matching posterior would not validate the shortcut, and keep exact fractions throughout. Do not introduce new urns, change the priors, or silently treat the unordered observation as two red draws.
The result
Let E denote the unordered observation that the two draws contain exactly one red and one blue. The urn priors are P(A)=1/2, P(B)=1/3, and P(C)=1/6, which sum to 1.
For draws without replacement, an urn with R red and B blue balls has
P(E|H)=RB/C(6,2)=RB/15,
since either color order is included.
| Urn | Prior | Likelihood P(E|H) | Weight | Posterior |
|---|---:|---:|---:|---:|
| A: (4R,2B) | 1/2 | 8/15 | 4/15 | 8/15 |
| B: (2R,4B) | 1/3 | 8/15 | 8/45 | 16/45 |
| C: (1R,5B) | 1/6 | 1/3 | 1/18 | 1/9 |
The evidence is the sum of the weights:
P(E)=4/15+8/45+1/18=(24+16+5)/90=1/2.
Thus the normalized posteriors are weight divided by 1/2, giving 8/15, 16/45, and 1/9. They normalize correctly:
8/15+16/45+1/9=(24+16+5)/45=1.
Because one red and one blue ball were removed, the remaining urns are A=(3R,1B), B=(1R,3B), and C=(0R,4B). Therefore the conditional probabilities that the third draw is red are respectively 3/4, 1/4, and 0. The posterior predictive probability is
P(three-red | E)=(8/15)(3/4)+(16/45)(1/4)+(1/9)(0)
=2/5+4/45=22/45.
For the tempting with-replacement shortcut, the unordered likelihood is
P_rep(E|H)=2(R/6)(B/6)=RB/18.
| Urn | Prior | Replacement likelihood | Weight | Posterior |
|---|---:|---:|---:|---:|
| A | 1/2 | 4/9 | 2/9 | 8/15 |
| B | 1/3 | 4/9 | 4/27 | 16/45 |
| C | 1/6 | 5/18 | 5/108 | 1/9 |
The replacement evidence is
P_rep(E)=2/9+4/27+5/108=(24+16+5)/108=5/12.
The replacement posterior again normalizes to
8/15+16/45+1/9=1.
With replacement, the third-draw red probabilities remain 2/3, 1/3, and 1/6. Hence
P_rep(three-red | E)=(8/15)(2/3)+(16/45)(1/3)+(1/9)(1/6)
=16/45+16/135+1/54=133/270.
Audit: the priors and normalized posteriors coincide in both models. The likelihoods, unnormalized weights, evidence probabilities, conditional third-draw probabilities, and predictive mixtures differ. Indeed, for every urn,
P_rep(E|H)=(5/6)P(E|H),
so every replacement weight is multiplied by the same factor 5/6. This common factor cancels during posterior normalization, explaining the coincident posteriors. It does not cancel in the evidence, changing 1/2 to 5/12. Moreover, replacement fails to remove the observed red and blue balls, so its third-draw conditional probabilities differ from 3/4, 1/4, and 0. Therefore the matching posterior is accidental scaling agreement and does not validate the replacement shortcut.
Time3 min 41 s
Cost$0.08
Finished2026-10-02